Problem of the Week #9 - July 30th, 2012

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Chris L T521

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Here is this week's problem!

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Problem: Let $f$ be function with measurable domain $D$. Show that $f$ is measurable if and only if the function $g$ defined on $\mathbb{R}$ by $g(x) = \begin{cases}f(x) & x\in D\\ 0 & x\notin D\end{cases}$ is measurable.

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Remember to read the POTW submission guidelines to find out how to submit your answers!

Chris L T521

Well-known member
Staff member
This week's question was correctly answered by girdav. Here's my solution.

Proof: Suppose $f$ is measurable and let $\alpha\in\mathbb{R}$. If $\alpha\geq 0$, then the set $\{x:g(x)>\alpha\}=\{x:f(x)>\alpha\}$, which is measurable. If $\alpha<0$, then $\{x:g(x)>\alpha\}=\{x:f(x)>\alpha\}\cup (\mathbb{R}\backslash D)$, which is again measurable. Conversely, suppose $g$ is measurable. Then $f=g|_{D}$, and since $D$ is measurable $f$ is measurable. Q.E.D.

Here's girdav's solution:

We can write $g(x)=f(x)\chi_D(x)$.

If $f$ is measurable, since the product of two measurable functions is measurable, so is $g$.

Conversely, assume $g$ measurable. Let $t\in\Bbb R$ and let $I_t:=(-\infty,t)$. If $t\leq 0$ then $f^{-1}(I_t)=g^{-1}(I_t)\cap D$ which is a measurable subset of $D$. If $t>0$, then $f^{-1}(I_t)=(g^{—1}(I_t)\cup D^c)\cap D=g^{-1}(I_t)\cap D$, which is a measurable subset of $D$.

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